Model: Let's simplify the model of a current in a wire, so we can be definite about what we are talking about. Take a wire (in the wire's frame) to have fixed positive charge density $\rho_{+}$ and assume the electrons at rest w.r.t the wire, with electron density $\rho_{-}$.
Introducing a current sets these electrons moving at some speed $v_\rm{drift}$ w.r.t wire, but leaves the positive charge fixed. We ask the following question:
What is the relationship between $\rho_{-}$ (the electron charge density at rest), and the electron density with current?
Answer: The density measured by the observer stationary w.r.t a current carrying wire is not the same as if the charges were stationary. They are related by a Lorentz transformation. Let's write the 4-current of the electrons when at rest, and when moving (with $c=1$):
$$J^\mu_\textrm{rest} = (\rho_{-},\vec{0})^\mu,\,\,\,\,\,J^\mu_\textrm{moving} = (\tilde{\rho},\vec{j})^\mu = {\Lambda(v)^\mu}_\nu J^\nu_\rm{rest}$$
where $\Lambda(v)$ is the Lorentz Boost between these two frames. Note in particular that $\boxed{\rho_{-} \neq \tilde{\rho}_{-}}$ because
$$J^2_\textrm{rest} = J^2_\textrm{moving}~~\implies ~~ \rho_{-}^2 = \tilde{\rho}^2_{-}-\vec{j}.\vec{j},$$
and $~\vec{j}\neq \vec{0}$.
This means that when you set up your problem, we have two possible scenarios:
$(i)$ $\rho_{-}+\rho_{+} = 0$, that is we ask that the electron density in the electrons rest frame has the same magnitude as the positive charge density in the stationary wire.
$(ii)$ $\tilde{\rho_{-}}+\rho_{+} = 0$, that is we ask that the electron density in the wire's rest frame has the same magnitude as the positive charge density in the stationary wire. This is the situation of zero force on a stationary external charge you talked about in your edit.
So the question you have to ask yourself, is what situation do you want to deal with? It seems that for the "explanation of magnetic force as a consequence of special relativity" you are interested in, one should consider case $(ii)$ as this allows you to see how a test charge, moving parallel to the wire with velocity $v$, experiencing a force due to a pure magnetic force in one frame (wire rest frame) $F = q v\times B$, is the same force experienced by the charge in its rest frame, effected only by the electric force, $F = q E$, in that frame (as in this frame it isn't moving).
I hope this helps. If you need further explanation, don't hesitate to ask.
Best Answer
You're completely correct to say in the reference frame of the electrons, the charge density of positive ions increases by a factor of $\gamma$. However if we're in the middle of the wire, there'll be a roughly equal amount of charge either side of the electrons, so no net force overcoming the electromotive force. Even if there was some net force on the electrons from these positive ions, it would be swamped by the force from the potential difference, otherwise we would have had no current to begin with. You could maybe even argue that whatever force is pulling the electric current along (e.g. a charged plate at the end of the wire) would be closer to the electrons by the same length contraction, so this might counter your increased force from a larger positive charge density.
Your other point about magnetic fields coming from moving charges is completely correct too. In the frame of the electrons they're not moving so just produce aan $\bf E$-field. In the lab frame, they're moving and making a $\bf B$-field. Relativity handles this by unifying $\bf E$ and $\bf B$ together in a field strength tensor, $F_{\mu\nu}$.